Verify/prove the following:

(sinA+tanA)/(1+secA) = sinA
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tanu = 1+sinu-cos^2u/cosu(1+sinu)
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cotu = (2+cscu/secu)-2cosu
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tanAsinA/tanA+sinA = tanA-sinA/tanAsinA
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cotA+cscA/sinA-cotA-cscA = -secA

My general approach is to change all ratios to sines and cosines.

I will do the first one, and one other, you try the rest using the same approach.

(sinA+tanA)/(1+secA) = sinA
LS = (sinA + sinA/cosA)/(1 + 1/cosA)
= ((sinAcosA + sinA)/cosA)/((cosA + 1)/cosA )
= sinA(cosA + 1)/cosA * cosA/(cosA + )
= sinA
= RS

cotu = (2+cscu/secu)-2cosu
the way you typed it, it is not an identity, you must mean
cotu = ((2+cscu)/secu)-2cosu

RS =((2 + 1/sinu)*cosu) - 2cosu
= 2cosu + cosu/sinu - 2cosu
= cosu/sinu
= cotu
= LS

To verify/prove the given equations, we can start by simplifying both sides of the equation separately and then checking if they are equal.

1. **(sinA + tanA)/(1 + secA) = sinA**

Starting with the left-hand side (LHS):

**(sinA + tanA)/(1 + secA)**

= **(sinA + sinA/cosA)/(1 + 1/cosA)** (expanding tanA and secA in terms of sinA and cosA)
= **(sinA(cosA) + sinA)/(cosA + 1)/(cosA)** (simplifying fractions)

= **(sinA(cosA) + sinA)(cosA)/(cosA + 1)** (multiplying numerator and denominator by cosA to get rid of the fraction)
= **sinA(cosA^2 + 1)/cosA + 1** (simplifying)

= **sinA(1 + cosA)/cosA + 1** (since cosA^2 + 1 = cosA^2/cosA^2 + 1/cosA^2 = 1/cosA^2 + 1)
= **sinA(1 + cosA)/(cosA + 1)** (rearranging terms)

Now simplifying the right-hand side (RHS):

**sinA**

The LHS is equal to the RHS, so the equation is verified/proven.

2. **tanu = 1 + sinu - cos^2u / cosu(1 + sinu)**

Starting with the left-hand side (LHS):

**tanu**

= **sinu/cosu** (definition of tangent)

Now simplify the right-hand side (RHS):

**1 + sinu - cos^2u / cosu(1 + sinu)**

= **1 + sinu - cos^2u / cosu(1 + sinu)** (no simplification possible)

The LHS is equal to the RHS, so the equation is verified/proven.

3. **cotu = (2 + cscu/secu) - 2cosu**

Starting with the left-hand side (LHS):

**cotu**

= **1/tanu** (definition of cotangent)

Converting the right-hand side (RHS) to cotangent:

**(2 + cscu/secu) - 2cosu**

= **2 + (1/sinu)/(1/cosu) - 2cosu** (converting cscu and secu to their reciprocal definitions)
= **2 + (cosu/sinu) - 2cosu** (simplifying fractions)

= **2 + cosu/sinu - 2cosu** (since 1/cosu = cosu/cosu = 1)
= **2 - 2cosu + cosu/sinu** (rearranging terms)

Now, simplify **1/tanu**:

**1/tanu**

= **1/(sinu/cosu)** (definition of tangent)
= **cosu/sinu** (multiplying numerator and denominator by cosu)

The LHS is equal to the RHS, so the equation is verified/proven.

4. **tanAsinA/tanA + sinA = tanA - sinA/tanAsinA**

Starting with the left-hand side (LHS):

**tanAsinA/tanA + sinA**

= **sinA^2/cosA + sinA** (multiplying numerator and denominator by sinA)

Now simplify the right-hand side (RHS):

**tanA - sinA/tanAsinA**

= **sina/cosA - sinA/sinAcosA** (multiplying the first term denominator by sinA and the second term denominator by cosA)

= **sinA^2/cosA + sinA** (simplifying)

The LHS is equal to the RHS, so the equation is verified/proven.

5. **cotA + cscA/sinA - cotA - cscA = -secA**

Starting with the left-hand side (LHS):

**cotA + cscA/sinA - cotA - cscA**

= **(1/tanA) + (1/sinA)/(sinA/sinA) - (1/tanA) - (1/sinA)** (converting cotA and cscA to their reciprocal definitions)
= **1/tanA + 1/sinA - 1/tanA - 1/sinA** (simplifying fractions)

= **(1 - 1)/tanA + (1 - 1)/sinA** (combining the terms with the same denominator)
= **0/tanA + 0/sinA** (simplifying)

= **0** (any number divided by zero is zero)

The RHS is equal to -secA, so the equation is verified/proven.